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a) MgO + 2HCl → MgCl2 + H2O (1)
b) \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
Theo PT1: \(n_{HCl}=2n_{MgO}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{10,95}{200}\times100\%=5,475\%\)
c) Theo PT1: \(n_{MgCl_2}=n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
\(\Sigma m_{dd}=6+200=206\left(g\right)\)
\(\Rightarrow C\%_{ddMgCl_2}=\dfrac{14,25}{206}\times100\%=6,92\%\)
d) HCl + KOH → KCl + H2O (2)
Theo PT2: \(n_{KOH}=n_{HCl}=0,3\left(mol\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(TheoPT:n_{AlCl_3}=\frac{2}{3}n_{H_2}=\frac{2}{3}.\frac{6,72}{22,4}=0,2\left(mol\right)\)
\(m_{ddsaup.ứ}=m_{Al}+m_{ddHCl}-m_{H_2}=0,2.27+\frac{2.0,3.36,5}{24,5\%}-0,3.2=94,19\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\frac{0,2.133,5}{94,19}=28,25\%\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(TheoPT:n_{NaOH}=n_{HCl}=0,6\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\frac{0,6.40}{40\%}=60\left(g\right)\)
\(\Rightarrow V_{NaOH}=\frac{60}{1,2}=50ml\)
1) K2CO3 +2HCl ------> 2KCl + CO2 +H2O
a) nHCl=\(\dfrac{200\cdot7,3\%}{36,5}\)=0,4 mol
Theo pthh nCO2= nK2CO3=1/2nHCl=0,2 mol
=>VCO2=0.2*22.4=4,48 (lít)
b)mdd K2CO3=\(\dfrac{0,2\cdot138}{20\%}\)=138(g)
c)mddA =mdd HCl +mdd K2CO3-mCO2=200+138-0.2*2=337,6
Theo pt nKCl=nHCl=0,4 mol
=>C%=\(\dfrac{0,4\cdot74,5}{337,6}\)=8,83%
2) 2Al +6HCl------> 2AlCl3 +3H2
a)nAl=0,2 mol
nHCl=\(\dfrac{365\cdot12\%}{36,5}\)=1,2 mol
Xét tỉ lệ \(\dfrac{n_{Al}}{2}\) < \(\dfrac{n_{HCl}}{6}\) => nAl hết nHCl dư, tính theo Al
Theo pt nAlCl3=nAl=0,2 mol
nH2=3/2nAl=0,3 mol
=>mAlCl3=133.5*0,2=26,7(g)
=>VH2=0,3*22,4=6,72 (l)
b)Theopt nHCl=3nAl=0,6 mol
mddA=mAl+mddHCl-mH2=5,4+365-0,3*2=369,8 (g)
=>C% HCl=\(\dfrac{\left(1,2-0,6\right)\cdot36,5\cdot100}{369,8}\)=5,92%
=>C%AlCl3=\(\dfrac{26,7\cdot100}{369,8}\)=7,22%
c) HCl+ KOH ---------> KCl+ H2O
Ta có nHCl dư=0,6 mol
Theo pt nKOH=nHCl=0,6
=>V=\(\dfrac{0,6}{1,5}\)=0,4 (l)
\(CaCO_3+2HCl-->CaCl_2+CO_2+H_2O\left(1\right)\)
0,25_______0,5_________0,25_____0,25
\(n_{CaCO_3}=0,25\left(mol\right)\)
=> \(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
\(m_{d^2HCl}=\frac{0,5.36,5.100}{14,6}=125\left(g\right)\)
\(m_{d^2sau}=25+125-0,25.44=139\left(g\right)\)
=> \(C\%_{CaCl_2}=\frac{0,25.111}{139}.100=19,96\%\)
a)\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=\frac{2,688}{22,4}=0,12\left(mol\right)\)
\(n_{SO^{ }_4}=\frac{1}{2}n_{H_2}=0,06\left(mol\right)\)
\(m_{muối}=m_{KL}+m_{SO_4}=6,44+0,06.96=12,2\left(g\right)\)
b) \(n_{H_2SO_4}=n_{H_2}=0,12\left(mol\right)\)
\(m_{ddH_2SO_4}=\frac{0,12.98}{9,8\%}=120\left(g\right)\)
Khí thoát ra là CO2
\(nCO_2=\frac{0,448}{22,4}=0,02\left(mol\right)\)
PTHH
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
0,02..........0,04.........0,02......................
\(C\%ddHCl=\frac{0,04\cdot36,5}{400}\cdot100\%=0,365\%\)
mdd sau phản ứng = \(0,02\cdot100+400-0,02\cdot44=401,12\left(g\right)\)
\(C\%ddspu=\frac{0,02\cdot111}{401,12}\cdot100\%\approx0,55\%\)
Kalicacbonat mà CaCo3 đâu ra vậy????